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QUESTION

# How do you find the roots of 12x^3+31x^2-17x-6=0?

Use the rational roots theorem, then factor to find:

x = 2/3" " or " "x = -1/4" " or " "x = -3

f(x) = 12x^3+31x^2-17x-6

By the rational roots theorem, any rational of f(x) are expressible in the form p/q for integers p, q with p a divisor of the constant term -6 and q a divisor of the coefficient 12 of the leading term.

That means that the only possible rational zeros are:

+-1/12, +-1/6, +-1/4, +-1/3, +-1/2, +-2/3, +-3/4, +-1, +-3/2, +-2, +-3, +-6

Notice that the signs of the coefficients of f(x) are in the pattern + + - -. By Descartes' Rule of Signs, since there is one change of sign, this cubic has exactly one positive Real zero.

So let's look among the positive possibilities:

f(1) = 12+31-17-6 = 20

f(1/2) = 12(1/8)+31(1/4)-17(1/2)-6 = (6+31-34-24)/4 = -21/4

f(2/3) = 12(8/27)+31(4/9)-17(2/3)-6 = (32+124-102-54)/9 = 0

So x=2/3 is a zero and (3x-2) a factor:

12x^3+31x^2-17x-6 = (3x-2)(4x^2+13x+3)

To factor the remaining quadratic, use an AC method:

Look for a pair of factors of AC=4*3=12 with sum B=13.

The pair 12, 1 works.

Use this pair to split the middle term and factor by grouping:

4x^2+13x+3 = (4x^2+12x)+(x+3)

color(white)(4x^2+13x+3) = 4x(x+3)+1(x+3)

color(white)(4x^2+13x+3) = (4x+1)(x+3)

So the other two zeros are:

x = -1/4" " and " "x = -3